By Rolle's Theorem, for a function $f:[a, b] \rightarrow R,$ if

a) $f$ is continuous on $[a, b]$

b) $f$ is continuous on $(a, b)$

c) $f(a)=f(b)$

Then, there exists some $c \in(a, b)$ such that $f^{\prime}(c)=0$

Therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

$f(x)=[x]$ for $x \in[5,9]$

It is evident that the given function $f(x)$ is not continuous at every integral point.

In particular, $f(x)$ is not continuous at $x=5$ and $x=9$

$\Rightarrow f(x)$ is not continuous in $[5,9]$

Also $f(5)=[5]=5$ and $f(9)=[9]=9$

$\therefore f(5) \neq f(9)$

The differentiability of $f$ in $(5,9)$ is checked as follows.

Let $\mathrm{n}$ be an integer such that $n \in(5,9)$

The left hand limit limit of $f$ at $x=n$ is.

$\mathop {\lim }\limits_{x \to 0'} \frac{{f(n + h) – f(n)}}{h} = \mathop {\lim }\limits_{x \to 0'} \frac{{[n + h] – [n]}}{h} = \mathop {\lim }\limits_{x \to 0'} \frac{{n – 1 – n}}{h} = \mathop {\lim }\limits_{x \to 0'} 0 = 0$

The right hand limit of $f$ at $\mathrm{x}=\mathrm{n}$ is,

$\mathop {\lim }\limits_{h \to {0^\prime }} \frac{{f(n + h) – f(n)}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{[n + h] – [n]}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{n – n}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} 0 = 0$

Since the left and right hand limits of $f$ at $x=n$ are not equal, $f$ is not differentiable at $x=n$

$\therefore f$ is not differentiable in $(5,9).$

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Rolle's Theorem.

Hence, Rolle's Theorem is not applicable for $f(x)=[x]$ for $x \in[5,9].$